Given the circle, midpoint $O(0,0)$ and radois $r$. Points $P(-r,0),Q(r,0)$ and $R(r\cos \alpha,r\sin \alpha),S(r\cos \beta,r\sin \beta)$. $\tan \theta=-\cot \frac{\alpha-\beta}{2}$. Point $T(\frac{r\cos \frac{\alpha+\beta}{2}}{\cos \frac{\alpha-\beta}{2}},\frac{r(\sin \frac{\alpha+\beta}{2}+\sin \frac{\alpha-\beta}{2})}{\cos \frac{\alpha-\beta}{2}})$. Area $\triangle PTQ\ =\ \frac{r^{2}(\sin \frac{\alpha+\beta}{2}+\sin \frac{\alpha-\beta}{2})}{\cos \frac{\alpha-\beta}{2}}$. Area $\triangle SRT\ =\ \frac{r^{2}(\sin \frac{\alpha+\beta}{2}+\sin \frac{\alpha-\beta}{2})}{\cos \frac{\alpha-\beta}{2}\sin^{2}\frac{\alpha-\beta}{2}}$. Ratio $\ =\ \frac{1}{\sin^{2}\frac{\alpha-\beta}{2}}=1+\cot^{2}\frac{\alpha-\beta}{2}=1+\tan^{2}\theta=\frac{1}{\cos^{2}\theta}$.

Rukevwe wrote: $PQ$ is a diameter of a circle. $PR$ and $QS$ are chords with intersection at $T$. If $\angle PTQ= \theta$, determine the ratio of the area of $\triangle QTP$ to the area of $\triangle SRT$ (i.e. area of $\triangle QTP$/area of $\triangle SRT$) in terms of trigonometric functions of $\theta$ First observe that $PQT\simeq SRT$ because $\angle TPQ=\angle RPQ=\angle RSQ=\angle RST$ and $\angle TQP=\angle SQP=\angle SRP=\angle SRT$. Hence if $\frac{TP}{TS}=r$, the desired ratio is $r^2$. Now observe that $\angle PST=\angle PSQ=90^\circ$, because $PQ$ is a diameter. Hence $\frac{TP}{TS}=\frac{1}{\cos (\angle STP)}=\frac{1}{\cos (180^\circ -\theta )}=\frac{-1}{\cos \theta}$ (this is positive because $\theta >90^\circ$). Hence the desired ratio is $\frac{1}{\cos ^2\theta}$.

Join SP, THEN, angle PSQ=90° angle PTQ= ∅ SO, angle SPR=∅-90° in ∆ STP, sin(∅-90°)=ST/PT - cos∅=ST/PT (PT/ ST)²=1/cos²∅ ∆PQT~∆SRT Ar of ∆PQT/ Ar of ∆ SRT=(PT/ ST)²=1/cos²∅ Ans. 1/cos²∅ @ Krishijivi